I have a large number of real, symmetric $3 \times 3$ matrices and would like to determine which one has the largest (most positive) eigenvalue, without calculating all of them individually for each matrix.

Is this possible? I.e do these matrices have some property whereby it’s possible to tell from their invariants which will have the largest eigenvalue? I suspect it isn’t directly, but wondered if it were possible to quickly come up with bounds on each one, and thereby exclude a (hopefully) large number of them from the eigenvalue calculation.

Any help or pointers would be very much appreciated.


1 Answer 1


I think you can try to derive the general form of det$(\pmb{A} - \lambda \pmb{I})$, with $\pmb{A}$ being a $3\times3$ matrix, in terms of determinants and traces.


$$ \pmb{A} = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} $$


$$ \pmb{X} = \begin{pmatrix} e & f \\ h & i \end{pmatrix}, \pmb{Y} = \begin{pmatrix} d & f \\ g & i \end{pmatrix}, \pmb{Z} = \begin{pmatrix} d & e \\ g & h \end{pmatrix} $$

then, making use of Laplace method: $$ \textrm{det}(\pmb{A} - \lambda \pmb{I}) = (a - \lambda) \cdot \textrm{det}(\pmb{X} - \lambda \pmb{I}) - b \cdot \textrm{det}(\pmb{Y} - \lambda \pmb{I}) + c \cdot \textrm{det}(\pmb{Z} - \lambda \pmb{I}) $$

Making all calculations you can arrive to the following generalization:

$\lambda^3 + \textrm{tr}(\pmb{A})\cdot\lambda^2 + (bd + cg - \textrm{det}(\pmb{X}) - a\cdot\textrm{tr}(\pmb{X})\cdot\lambda -b\cdot\textrm{det}(\pmb{Y}) + c\cdot\textrm{det}(\pmb{Z}) - afh$

Probably you can further simplify the polynomial with the substitution $\lambda = t - \frac{\textrm{tr}(\pmb{A})}{3}$ (see here)

In this way you might only need to calculate traces, determinants and simple multiplications to estimate the largest positive eigenvalue. If not enough, I hope this will at least give you some ideas on how to continue. Good luck!


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